Calculating Weight in an Elevator

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SUMMARY

The discussion centers on calculating the weight of a person with a mass of 71 kg in an elevator, using a velocity vs. time graph. The graph indicates varying accelerations: 4 m/s² from 0 to 2 seconds, 0 m/s² from 2 to 6 seconds, and -2 m/s² from 6 to 10 seconds. The key takeaway is that the reading on a spring scale does not equal the person's weight due to the elevator's acceleration. Instead, the weight is defined as the force of gravity acting on the mass, calculated as mg, where g is approximately 9.81 m/s².

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Familiarity with free body diagrams
  • Knowledge of basic kinematics and acceleration
  • Ability to interpret velocity vs. time graphs
NEXT STEPS
  • Study the concept of weight vs. apparent weight in non-inertial frames
  • Learn how to construct and analyze free body diagrams
  • Explore the relationship between acceleration and forces in elevators
  • Review the principles of kinematics related to velocity and acceleration
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of forces in varying acceleration scenarios, particularly in the context of elevators.

rysezhae
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Homework Statement



I need to find the weight of a person who has mass 71 kg, and is riding an elevator, at different time intervals. The intervals are 1,5, and 9 seconds. I was given a velocity vs time graph, so I'll try to describe as best i can: The graph starts at (0,0), the slope is 4/1, however when it reaches (2,8)- 2 seconds and 8 m/s - it flatlines for about 4 seconds, then the slope changes to -2/1 and stops when it reaches (0,10) - 10 seconds and 0 m/s.

Homework Equations



No relevant equations

The Attempt at a Solution



I tried to find the acceleration, which I know is the slope of the graph, then multiplying it by the mass, but obviously that didnt work, also my answer has to have two sig figs, and of my course my attempted answers give me 3 or 4. I think if i could form an equation for the graph and then find the differential, then substitute in the time intervals, that would give me a better number to multiply the mass by
 
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Hi rysezhae. Welcome to Physics Forums.

Let me guess. You are trying to find the reading that a spring scale would register with a mass of 71 kg sitting on the scale. This is not always the person's weight. His weight is defined as the force that gravity exerts on him, mg, and this doesn't change even though the reading on the spring scale may vary with the acceleration of the elevator. If the acceleration of the elevator equals zero, the spring scale will register his actual weight. So the reading registered by the spring scale is not just the mass times the acceleration; otherwise, the scale would register zero if the elevator were not accelerating.

Back to your acceleration calculations. If I understand you correctly, the acceleration is 4m/s^2 from 0 to 2 seconds, 0 m/s^2 from 2 to 6 seconds, and -2 m/s^2 from 6 seconds to 10 seconds. Is that correct. Now focus on the person as a free body, and identify the two forces acting on him. Then write down (symbolically) the Newton's 2nd law force balance on this free body. This should give you enough to get what you need.
 

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